What is the extraneous solution to these equations? $\dfrac{x^2 + x}{x + 3} = \dfrac{30}{x + 3}$
Multiply both sides by $x + 3$ $ \dfrac{x^2 + x}{x + 3} (x + 3) = \dfrac{30}{x + 3} (x + 3)$ $ x^2 + x = 30$ Subtract $30$ from both sides: $ x^2 + x - (30) = 30 - (30)$ $ x^2 + x - 30 = 0$ Factor the expression: $ (x + 6)(x - 5) = 0$ Therefore $x = -6$ or $x = 5$ The original expression is defined at $x = -6$ and $x = 5$, so there are no extraneous solutions.